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3 years ago

Solve: x^+4/x-1 = 5/x-1

1 answer:

7 0

The first step for solving this equation is to determine the defined range.
$\frac{ x^{4} }{x-1} = \frac{5}{x-1}$ , x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
$x^{4}$ = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- $\sqrt[4]{5}$
Separate the solutions.
x = $\sqrt[4]{5}$ , x ≠ 1
x = - $\sqrt[4]{5}$
Check if the solution is in the defined range.
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
This means that the final solution to your question are the following:
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
Let me know if you have any further questions.
:)

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ABCD is a parallelogram. If AB =3x+4, and CD =4x-1, find the length of AB

RSB [31]

If AB = CD, then 3x+4 is equal to 4x-1

3x + 4 = 4x - 1
Add one to both sides to cancel out.
3x + 5 = 4x
Subtract 3x from both sides to cancel out.
5 = x

Now, substitute the 5 in place of the x in the equation for AB (3x+4)

3(5) + 4
15 + 4
19

Final Answer: B) 19

7 0

4 years ago

Read 2 more answers

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$